Estonian Mathematical Olympiad

Let $P$ be the center of the square pasture and let some scarecrow be at point $O$. Since the area bounded by the scarecrows must entirely contain the interior of the square and the points of the square are at least distance $1$ from the center of the square, we have $PO\ge 1+1=2$ with the equality holding only if the midpoint of $PO$ coincides with the midpoint of one of the sides of the square. Consider first the case where $PO=2$. Let $A$ and $B$ be points on the unit circle centered at $O$ such that $PA$ and $PB$ are tangents to this circle, and let $C$ be the midpoint of segment $PO$ (Fig. 39). Since the tangents $PA$ and $PB$ are perpendicular to the radii $OA$ and $OB$ at the points of tangency, the points $P,A,O,B$ lie on a circle with diameter $PO$. Thus, $OA=OB=OC=PC=AC=BC=1$, and since triangles $OAC$ and $OBC$ are equilateral, $\angle AOB=60^{\circ }+60^{\circ }=120^{\circ }$ and $\angle APB=180^{\circ }-120^{\circ }=60^{\circ }$.

Fig. 39 Fig. 40

Hence, a scarecrow covers an angle of $60^{\circ }$ of the horizon as viewed from the center of the pasture. If $PO>2$, the covered angle is obviously smaller. Since the scarecrows must cover the entire $360^{\circ }$ horizon as seen from $P$, at least $6$ scarecrows are needed. However, covering this area with exactly $6$ scarecrows would only be possible if each scarecrow covers exactly $60^{\circ }$. This would require all scarecrows to be at a distance of $2$ from the center of the pasture. This, in turn, would be possible only if the midpoint of each segment connecting the scarecrow to the center of the square is at the midpoint of one of the sides of the square. Since the square has only $4$ sides, it is impossible to place $6$ scarecrows this way. Therefore, at least $7$ scarecrows are needed.

We will show that $7$ scarecrows are sufficient. Let the coordinate origin be the center of the pasture and let the coordinate axes be parallel to the sides of the pasture. The scarecrows can be placed, for example, at the points $O_1=(2,1)$, $O_2=(2-\sqrt{3},2)$, $O_3=(-1.6,1.8)$, $O_4=(-2,0)$, $O_5=(-1.6,-1.8)$, $O_6=(2-\sqrt{3},-2)$ and $O_7=(2,-1)$ (Fig. 40). We will prove this.

First, it is clear that the areas affected by the scarecrows at points $O_1,O_2,O_4,O_6$, and $O_7$ touch the square but do not overlap with it. Since the distance of point $O_3$ from the upper left corner of the square is $1$, and it is even further from the other points of the square, the influence area of the scarecrow at point $O_3$ does not overlap with the pasture. The same holds for $O_5$.

Secondly, we show that the scarecrows cover the entire $360^{\circ }$ horizon as seen from the pasture. For this, we show that the length of each side of the polygon $O_1{O}_2...O_7$ is at most $2$. Clearly, $|O_7{O}_1|=2$ and direct inspection shows that $O_1{O}_2=O_6{O}_7=2$. It remains to see that $$\begin{array}{rl}(O_2{O}_3{)}^2& =(2-\sqrt{3}+1.6{)}^2+(2-1.8{)}^2=(3.6-\sqrt{3}{)}^2+0.2^2\\ & <(3.6-\sqrt{2.89}{)}^2+0.2^2=(3.6-1.7{)}^2+0.2^2\\ & =1.9^2+0.2^2=3.61+0.04<4,\\ (O_3{O}_4{)}^2& =(-1.6+2{)}^2+1.8^2=0.4^2+1.8^2=0.16+3.24=3.4<4.\end{array}$$ Thus $O_2{O}_3<2$ and $O_3{O}_4<2$; by symmetry $O_4{O}_5<2$ and $O_5{O}_6<2$. Therefore, $7$ scarecrows are sufficient to confine the goat.